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Nuclear fission produces quark mass

  1. Jun 5, 2004 #1
    "nuclear" fission produces quark mass

    A top quark is roughly 32 times more massive than a bottom quark.
    A charm quark is roughly 8 times more massive than a strange quark.
    And a down quark is roughly 2 times more massive than an up quark.

    In nuclear fission 2 neutrons yield 4 neutrons which become 8 neutrons
    16 then 32 etc.

    So how are quark masses produced?
    If an up quark is made from two smaller particles and separating those
    particles creates other quarks (quark confinement analogy)successively
    heavier quarks will be created if the force holding the two particles
    of the up quark together gets stronger with increasing separation.
    Suppose that the space between the two particles of the up quark is
    occupied by
    groups of unstable particles of another kind ( analagous to unstable
    atomic nuclei), and one of the particles that constitute the up quark
    emits the equivalent of a neutron.An exponential cascade will be set
    in motion yielding
    more and more force carriers as distance increases.This is why the
    force between the two particles of the up quark gets stronger with
    increasing distance.
    But why are the numbers 4 and 16 missing from the cascade - why are no
    quarks produced at the distances corresponding to these
    quarks are made - and they go by the name of Dark Energy!
    There are thus 4 quark families ( a missing lepton too presumably).
  2. jcsd
  3. Jun 7, 2004 #2
    My answer to this is the properties of directional invariance. Imagine a cube with its 8 vertices, if one vertex is removed, the cube cease to exist. Each vertex is the intersection of three directions. If each vertex is considered as one directional property then there are eight such property for a cube to exist. These are the eight properties of directional invariance. top-right-front, top-right-back, down-right-front, down-right-back, top-left-front, top-left-back, down-left-front, down-left-back.

    The generation of family seems to follow odd power of two's. [itex]2^1, 2^3, 2^5, 2^7[/itex] where [itex] 2^7[/itex] is the 4th family.

    If four vertices are removed from the cube then another stable configuration is formed called a tetrahedron. All stable configurations seems to satisfy Euler formula for polyhedra [itex] V - E + F = 2[/itex] where v is the number of vertices, E is the number of edges, F is the number of faces.
    Last edited: Jun 7, 2004
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